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College Algebra: Solving Rational Inequalities


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About this Lesson

  • Type: Video Tutorial
  • Length: 8:42
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 94 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Equations & Inequalities (50 lessons, $65.34)
College Algebra: Inequalities: Rationals, Radicals (3 lessons, $4.95)

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Recent Reviews

relearning fractions
~ wmartinez

This piece was great for recalling my old algebra.

relearning fractions
~ wmartinez

This piece was great for recalling my old algebra.

So, when you're solving a quadratic inequality you remember that you first have to factor. So, you have those two things multiplying together and then you can figure out the sign of each of those things with the little number, sort of sign chart thing, and figure out where the inequality answer lies. You know, the same exact method, at least in theory, can be used--if you don't have just certainly a quadratic, but if you have a rational inequality. So, that is a fraction with an inequality. Let me actually show you this by an example. Suppose that I wanted to find all the values for x, so that 3x + 4 all divided by x + 1, all that value is going to be always either less than or equal to 2. So, I want to find all the x's that I can plug in here so that this thing is always true.
Well, how do you proceed? Well, the first thing I want to always do, just like you would do with a quadratic, is to get everything over and have something on one side and then 0 on the other. So, we always want to have things compared to 0, because then we know its either positive or negative and then we know whether its plus or minus or minus plus or plus plus or minus minus and so on. But with a 2 here, that's not good. So, the first step always is to bring everything over and produce a 0 on one side. So, what I'll do is I'll actually bring this over to the other side and have the 0 on the right. Let me do this here for you. So, this would look like 3x + 4 divided by x +1, but now I'm going to subtract a 2 from both sides. So, I've got a -2 and that's less than or equal to 0. So, there's the happy 0 I want, that's great.
But now I've got all this stuff here. What I'll do actually is get a common denominator and I'll combine this fraction with this fraction. So, it's going to be still a little bit more Algebra we have to do here. I have to get a common bottom. Remember this 2 is actually 2 over 1, if you want to think of it that way. So, I have to multiply top and bottom by something. In this case it's going to be x + 1. So, to get a common denominator I'm going to just keep the first guy, 3x + 4 over x + 1, but to add or subtract fractions I need a common bottom. So, I'm going to do this, (x + 1) (x + 1) and that's less than or equal to 0. So, now I have the common bottom, so I can just subtract the tops. So, I'm going to subtract the tops now and what would I get?
Well, I'll write this out here now. What I would see is 3x + 4 - 2(x + 1), that's all on the top and on the bottom I have that common bottom, x + 1, and that's all less than or equal to 0. So, that's where we are right now. I brought the 2 over, got a common denominator, combined it. Let's try to actually clean up that top there. Now, remember what I've got to do. I've got to now distribute and I've got to distribute not just the 2, but the -2. I've got to subtract all this. Don't make classic mistake number 4, the subtracting mistake. You have to share the negativity. So, this whole thing has to be distributed, not only to the x, but also to the plus 1. That's a great, great mistake that you should never make.
So, let's see what we'd have here. I'd have a 3x and then I subtract 2x's away, that just leaves me with x. So, this would just be x and then here I see a 4, but then I have a -2 times 1. Which is a -2. So, I have 4 - 2. Well, that's just plus 2. On the bottom I have x + 1. So, actually this cleaned up quite nicely. You have to admit that after you take this thing and cleaned it up it came out quite nice. That happens all the time. You ever have something that looks sort of dingy, but then you clean up. It looks brand new. Right? This is great. Exact same thing here, this is a math example. Now what do you do now?
Well, now I'm going to use the exact same method that I used for solving the quadratic inequalities. Namely, I'm going to ask when things are 0. But I have to be a little bit careful here because I'm going to ask when the numerator is 0, and that's okay. But I'm also going to ask when the denominator is 0 and that's of course no man's land. So, we have to be really careful with this. So, I'm going to take a number line and I'm going to mark down the regions where the top is 0 and regions where the bottom is 0. So, the top is 0, let's see, when is that equal to 0? Well, when x = -2. So, I'm going to put in a -2 right here and I'm going to write a big 0 there. You could even, if you want, put like a dotted line to show that it's going to separate this world into two pieces right here.
But now I'm also going to put down where the bottom is 0. Now, where is the bottom equal to 0? Well, you can see that happens when x = -1. If x = -1, I have a 0 in the denominator. So, I mark that down, but I'm going to put like a U here. I'm not going to put a number because we know the fraction is undefined. We know that a fraction is undefined when the bottom is 0. So, I'm going to put a U there, but I'm going to mark that down anyway, because in fact it may divide the world up even more. So, notice what I'm doing. It's the exact same idea we did with the quadratics.
I find out where each piece is equal to 0. Now, with the quadratic they were factors. Now, with fractions that are on top of each other, so its division rather than multiplication, but the idea is the same. I'm going to put down where they're 0. However, where the bottom is 0, I put an undefined, because I know this thing is not defined there. Now, I just proceed as normal. I want to find out what the sign of this thing is on these different regions. So, what do I do? All I've got to do is pick a point in there and plug in and see what this produces, a positive or a negative. Whatever it produces at that point, its going to produce the same thing at all the points in this region.
So, let's see. Let's pick a point way out here to the right of -1. How about 0? So, when we plug in x = 0, into this thing and see what it produces. If I put a 0 in here and here, I see 2 over 1. Well, I don't care that its 2, all I care about is its sign and I see it's positive. So, if it's positive at this point it's going to be positive everywhere here. So, this is all positive land. Now, I need to pick a point that's between -1 and -2. Well, there aren't that many easy points. There's like no integers in there, for example. So, I'll have to pick something like -1.5. That's something between -1 and -2. Sorry about that. So, -1.5 is what we'll pick. If you put in -1.5 and add 2, what do you get? Well, 2 - 1.5 is a .5, so it's positive. All I care about is the sign, it's positive.
What about when I put in a -1.5 here? If I have -1.5 and I add 1, I'm left with -.5. So, I have a positive on top and a negative on the bottom. When I divide them, that result is negative. So, this land is actually negative land. What happens here? Well, here I could pick something like -10. If I put in -10 here, I'd see -10 + 2. That's negative. -10 +1, that's negative. Do you see how I'm not actually computing the values, but just thinking through the sign? Like -10 + 2 is plainly negative. This is plainly negative. Negative divided by a negative, positive again. So, this land is all positive.
So, now I can look at my chart and answer the question. The question was, "When is this thing less than or equal to 0?" So, where is it negative? Well, it's negative right in this region right here. It's negative all in there. But what about the end points? I'm allowed to equal 0. So, do I include the end points or not? What do you think? Well, here at -2 I know this thing equals 0, because the top is 0. So, I want to include that. So, I put in my bracket like that. I should include -1 because that makes it 0 too. Right? Wrong. That is wrong, because if I plug in -1, it makes the bottom 0 and thus makes the whole thing undefined. So, this can never be in the solution because I'm not even allowed to plug that in.
So, I have to put one of these open parentheses just to graze it like that. So, there's the solution graphically. If you want to write it out in interval notation, it would be [-2, -1) and that's the interval notation answer. That means, that if you pick any x value in this region and plug it in here this will be a true inequality. But if you pick anything else out in this part of the world, this will no longer be true. So, it's the exact same thing as when you have products within inequalities. You have to look at where the tops and bottoms are now 0, but carefully mark the denominator part as undefined and then just make your sign chart, see what you want, report the news. I'll take a look at another one up next. I'll see you there.
Equations and Inequalities
Inequalities - Rationals and Radicals
Solving Rational Inequalities Page [1 of 2]

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